0

I have a query: Let us  have 2 particles and 2 corresponding wavefunctions, under two incompatible Hamiltonians ( $H_1,H_2$).

$$\Psi_1(x_1,t_1)= e^{-x_1} e^{i\sin(\pi t_1/3)}+e^{-x_1^2}e^{i\cos(\pi t_1/5)}$$ and $$\Psi_2(x_2,t_2)= e^{-x_2}e^{i\cos(\pi t_2/12)} +e^{-x_2^2}e^{i\sin(3\pi t_2/20)}$$ These are such that they become the same expression at $t_1= 1\,\text s$ and $t_2=2\,\text s$ respectively.

Now, after $t_1=1\, \text s$ , if we act the Hamiltonian $H_2$ on particle 1 , what can we expect to happen?

SX849
  • 159
  • This is all going to make way, way more sense to you if you learn about "second quantization". The usual thing of symmetrizing and anti-symmetrizing wave functions is very stupid and IMHO should only be taught after second quantization. See for example this post. – DanielSank Nov 27 '21 at 04:45

1 Answers1

0

The comments clarified your question a bit.

This question has nothing to do with entanglement.

  • A two particle wave function is $\Psi(x_1,x_2,t)$. None of your function is of that form. They are one particle wave functions.

  • The two particle wave function describes entangled particles when it cannot be written in the form $$ \Psi(x_1,x_2,t)=\Psi_1(x_1,t)\Psi_2(x_2,t)\,. $$

To the idea you had with your question: It is conceivable that you can smoothly paste together "two" wave functions $\Psi_1(x,t)$ and $\Psi_2(x,t)$ at some fixed $t$ but this will still give a single particle wave function because its space argument is one dimensional.

If you like here is my first version of the answer:

Doing calculations such as

\begin{align} \frac{\partial\Psi_1}{\partial t_1}&=e^{-x_1}e^{i\sin(\pi t_1/3)}\cos(\pi t_1/3)\frac{i\pi}{3}-e^{-x_1^2}e^{i\cos(\pi t_1/5)}\sin(\pi t_1/5)\frac{i\pi}{5}\,,\\ \frac{\partial\Psi_1}{\partial x_1}&=-e^{-x_1}e^{i\sin(\pi t_1/3)}-2x_1e^{-x_1^2}e^{i\cos(\pi t_1/5)}\,,\\ \frac{\partial^2\Psi_1}{\partial x_1^2}&=e^{-x_1}e^{i\sin(\pi t_1/3)}-2e^{-x_1^2}e^{i\cos(\pi t_1/5)}+4x_1^2e^{-x_1^2}e^{i\cos(\pi t_1/5)}\,,\\ \end{align} -and same for $\Psi_2$- one should be able to write down the Hamiltonians $H_1,H_2$ explicitly so that each $\Psi_1$ and $\Psi_2$ satisfies the Schrödinger equation $$ i\hbar\frac{\partial \Psi_i}{\partial t_i}=H_i\Psi_i\,. $$ Regarding your question what happens when we act $H_2$ on $\Psi_1\,:$ It is unlikely that $$ i\hbar\frac{\partial \Psi_1}{\partial t_1}=H_2\Psi_i $$ holds. In other words, it is unlikely that $H_2$ is the energy operator of "particle 1". The fact that $\Psi_1(\,.,t_1)$ and $\Psi_2(\,.,t_2)$ agree for two $t_1$ and $t_2$ is obviously totally irrelevant because the time evolution of $\Psi_1$ is governed by $H_1$ and not by $H_2\,.$

Kurt G.
  • 1,789
  • Let me add some more clarity to the question: At t=1, we totally switch off Hamiltonian $H_1$ , and act $H_2$ on the state of particle 1 at t=1. Since the Hamiltonians are incompatible, the particle's memory of being under $H_1$ will have disappeared. Also, the common state that we get satisfies also the Schrodinger equation created by $H_2$ so we may take this state as an initial state of particle 1, and then time evolve it via $H_2$. What will happen then? – SX849 Nov 26 '21 at 15:09
  • I'd buy that if you ensure that you could smoothly paste together $\Psi_1$ and $\Psi_2$ at, say $t=1,.$ Even if your functions above don't do that it is conceivable that one can find "two" such functions. I wrote "two" in quotation marks because you will have a single state of a single particle. I don't get what this has to do with entanglement, or even with two particles. – Kurt G. Nov 26 '21 at 18:33
  • Alright, I understand that it perhaps has nothing to do with entanglement. Thanks for clearing this! – SX849 Nov 26 '21 at 21:01
  • As long as the particles live on a compact space (so their positions are bounded), an expression of the form $\Psi =\psi_1(x_1) + \psi_2(x_2)$ is a perfectly valid 2-particle wavefunction because $\Psi=\psi_1(x_1)\mathbf 1(x_2) + \mathbf 1(x_1) \psi_2(x_2)$, where $\mathbf 1(x) = 1$. Such a state is necessarily entangled; perhaps that's what the OP was referring to. – J. Murray Nov 27 '21 at 05:19
  • @J.Murray, Mr. Kurt G. has said that the two states are not entangled because they are not in the necessary entangled form. Could you please explain that, as you have said that the state is entangled. – SX849 Nov 27 '21 at 07:12